Introduction. Formulating LP Problems LEARNING OBJECTIVES. Requirements of a Linear Programming Problem. LP Properties and Assumptions

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1 Valua%on and pricing (November 5, 2013) LEARNING OBJETIVES Lecture 10 Linear Programming (part 1) Olivier J. de Jong, LL.M., MM., MBA, FD, FFA, AA 1. Understand the basic assumptions and properties of linear programming (LP). 2. Graphically solve any LP problem that has only two variables by both the corner point and isoprofit line methods. 3. Understand special issues in LP such as infeasibility, unboundedness, redundancy, and alternative optimal solutions. 4. Understand the role of sensitivity analysis. 5. Use Excel spreadsheets to solve LP problems. opyright 2015 Pearson Education, Inc. 7 2 Introduction Many management decisions involve making the most effective use of limited resources Linear programming (LP) Widely used mathematical modeling technique Planning and decision making relative to resource allocation Broader field of mathematical programming Here programming refers to modeling and solving a problem mathematically Requirements of a Linear Programming Problem Four properties in common Seek to maximize or minimize some quantity (the objective function) Restrictions or s are present Alternative courses of action are available Linear equations or inequalities opyright 2015 Pearson Education, Inc. 7 3 opyright 2015 Pearson Education, Inc. 7 4 LP Properties and Assumptions TABLE 7.1 PROPERTIES OF LINEAR PROGRAMS 1. One objective function 2. One or more s 3. Alternative courses of action 4. Objective function and s are linear proportionality and divisibility 5. ertainty 6. Divisibility 7. Nonnegative variables Formulating LP Problems Developing a mathematical model to represent the managerial problem Steps in formulating a LP problem 1. ompletely understand the managerial problem being faced 2. Identify the objective and the s 3. Define the decision variables 4. Use the decision variables to write mathematical expressions for the objective function and the s opyright 2015 Pearson Education, Inc. 7 5 opyright 2015 Pearson Education, Inc. 7 6

2 Formulating LP Problems ommon LP application product mix problem Two or more products are produced using limited resources Maximize profit based on the profit contribution per unit of each product Determine how many units of each product to produce Flair Furniture ompany Flair Furniture produces inexpensive tables and chairs Processes are similar, both require carpentry work and painting and varnishing Each table takes 4 hours of carpentry and 2 hours of painting and varnishing Each chair requires 3 of carpentry and 1 hour of painting and varnishing There are 240 hours of carpentry time available and 100 hours of painting and varnishing Each table yields a profit of $70 and each chair a profit of $50 opyright 2015 Pearson Education, Inc. 7 7 opyright 2015 Pearson Education, Inc. 7 8 Flair Furniture ompany The company wants to determine the best combination of tables and chairs to produce to reach the maximum profit TABLE 7.2 DEPARTMENT HOURS REQUIRED TO PRODUE 1 UNIT (T) TABLES () HAIRS AVAILABLE HOURS THIS WEEK arpentry Painting and varnishing Profit per unit $70 $50 Flair Furniture ompany The objective is Maximize profit The s are The hours of carpentry time used cannot exceed 240 hours per week The hours of painting and varnishing time used cannot exceed 100 hours per week The decision variables are T = number of tables to be produced per week = number of chairs to be produced per week opyright 2015 Pearson Education, Inc. 7 9 opyright 2015 Pearson Education, Inc Flair Furniture ompany reate objective function in terms of T and Maximize profit = $70T + $50 Develop mathematical relationships for the two s For carpentry, total time used is (4 hours per table)(number of tables produced) + (3 hours per chair)(number of chairs produced) First is arpentry time used arpentry time available 4T (hours of carpentry time) Flair Furniture ompany Similarly Painting and varnishing time used Painting and varnishing time available 2 T (hours of painting and varnishing time) This means that each table produced requires two hours of painting and varnishing time Both of these s restrict production capacity and affect total profit opyright 2015 Pearson Education, Inc opyright 2015 Pearson Education, Inc. 7 12

3 Flair Furniture ompany The values for T and must be nonnegative T 0 (number of tables produced is greater than or equal to 0) 0 (number of chairs produced is greater than or equal to 0) The complete problem stated mathematically subject to Maximize profit = $70T + $50 4T (carpentry ) 2T (painting and varnishing ) T, 0 (nonnegativity ) opyright 2015 Pearson Education, Inc Graphical Solution to an LP Problem Easiest way to solve a small LP problems is graphically Only works when there are just two decision variables Not possible to plot a solution for more than two variables Provides valuable insight into how other approaches work Nonnegativity s mean that we are always working in the first (or northeast) quadrant of a graph opyright 2015 Pearson Education, Inc FIGURE 7.1 Quadrant ontaining All Positive Values This Axis Represents the onstraint T 0 This Axis Represents the onstraint 0 opyright 2015 Pearson Education, Inc The first step is to identify a set or region of feasible solutions Plot each equation on a graph Graph the equality portion of the equations 4T + 3 = 240 Solve for the axis intercepts and draw the line opyright 2015 Pearson Education, Inc When Flair produces no tables, the carpentry is: 4(0) + 3 = = 240 = 80 Similarly for no chairs: 4T + 3(0) = 240 4T = 240 T = 60 This line is shown on the following graph opyright 2015 Pearson Education, Inc FIGURE 7.2 Graph of arpentry onstraint Equation (T = 0, = 80) (T = 60, = 0) opyright 2015 Pearson Education, Inc. 7 18

4 FIGURE 7.3 Region that Satisfies the arpentry onstraint (30, 40) Any point on or below the plot will not violate the restriction Any point above the plot will violate the restriction (70, 40) 20 (30, 20) opyright 2015 Pearson Education, Inc The point (30, 40) lies on the line and exactly satisfies the 4(30) + 3(40) = 240 The point (30, 20) lies below the line and satisfies the 4(30) + 3(20) = 180 The point (70, 40) lies above the line and does not satisfy the 4(70) + 3(40) = 400 opyright 2015 Pearson Education, Inc FIGURE 7.4 Region that Satisfies the Painting and Varnishing onstraint (T = 0, = 100) (T = 50, = 0) opyright 2015 Pearson Education, Inc To produce tables and chairs, both departments must be used Find a solution that satisfies both s simultaneously A new graph shows both plots The feasible region is where all s are satisfied Any point inside this region is a feasible solution Any point outside the region is an infeasible solution opyright 2015 Pearson Education, Inc FIGURE 7.5 Feasible Solution Region Painting/Varnishing onstraint 20 Feasible arpentry onstraint Region opyright 2015 Pearson Education, Inc For the point (30, 20) arpentry Painting For the point (70, 40) arpentry Painting 4T hours available (4)(30) + (3)(20) = 180 hours used 2T hours available (2)(30) + (1)(20) = 80 hours used 4T hours available (4)(70) + (3)(40) = 400 hours used 2T hours available (2)(70) + (1)(40) = 180 hours used opyright 2015 Pearson Education, Inc ü ü û û

5 For the point (50, 5) arpentry Painting 4T hours available (4)(50) + (3)(5) = 215 hours used 2T hours available (2)(50) + (1)(5) = 105 hours used ü û Find the optimal solution from the many possible solutions Speediest method is to use the isoprofit line Starting with a small possible profit value, graph the objective function Move the objective function line in the direction of increasing profit while maintaining the slope The last point it touches in the feasible region is the optimal solution opyright 2015 Pearson Education, Inc opyright 2015 Pearson Education, Inc hoose a profit of $2,100 The objective function is $2,100 = 70T + 50 Solving for the axis intercepts, draw the graph Obviously not the best possible solution Further graphs can be created using larger profits The further we move from the origin, the larger the profit The highest profit ($4,100) will be generated when the isoprofit line passes through the point (30, 40) opyright 2015 Pearson Education, Inc FIGURE 7.6 Profit line of $2, $2,100 = $70T + $50 (30, 0) opyright 2015 Pearson Education, Inc FIGURE 7.7 Four Isoprofit Lines $3,500 = $70T + $50 $2,800 = $70T + $50 $2,100 = $70T + $50 $4,200 = $70T + $50 opyright 2015 Pearson Education, Inc FIGURE 7.8 Optimal Solution Maximum Profit Line Optimal Solution Point (T = 30, = 40) $4,100 = $70T + $50 opyright 2015 Pearson Education, Inc. 7 30

6 orner Point Solution Method The corner point method for solving LP problems Look at the profit at every corner point of the feasible region Mathematical theory is that an optimal solution must lie at one of the corner points or extreme points orner Point Solution Method FIGURE 7.9 Four orner Points of the Feasible Region 100 (0, 80) (?) (50, 0) (0, 0) opyright 2015 Pearson Education, Inc opyright 2015 Pearson Education, Inc orner Point Solution Method Solve for the intersection of the two lines Using the elimination method to solve simultaneous equations method, select a variable to be eliminated Eliminate T by multiplying the second equation by 2 and add it to the first equation 2(2T + 1 = 100) = 4T 2 = 200 4T + 3 = 240 (carpentry) 4T 2 = 200 (painting) = 40 opyright 2015 Pearson Education, Inc orner Point Solution Method Substitute = 40 into either equation to solve for T 4T + 3(40) = 240 4T = 240 4T = 120 T = 30 TABLE 7.3 Feasible orner Points and Profits Thus the corner point is (30, 40) NUMBER OF TABLES (T) NUMBER OF HAIRS () PROFIT = $70T + $ $ $3, $4, $4,100 Highest profit Optimal Solution opyright 2015 Pearson Education, Inc Slack and Surplus Slack is the amount of a resource that is not used For a less-than-or-equal Slack = (Amount of resource available) (Amount of resource used) Flair decides to produce 20 tables and 25 chairs 4(20) + 3(25) = 155 (carpentry time used) 240 = (carpentry time available) = 85 (Slack time in carpentry) Slack and Surplus At the optimal solution, slack is 0 as all 240 Slack is the amount of a resource hours are used that is not used For a less-than-or-equal Slack = (Amount of resource available) (Amount of resource used) Flair decides to produce 20 tables and 25 chairs 4(20) + 3(25) = 155 (carpentry time used) 240 = (carpentry time available) = 85 (Slack time in carpentry) opyright 2015 Pearson Education, Inc opyright 2015 Pearson Education, Inc. 7 36

7 Slack and Surplus Surplus is used with a greater-than-or-equalto to indicate the amount by which the right-hand side of the is exceeded Surplus = (Actual amount) (Minimum amount) New T + 42 If T = 20 and = 25, then = 45 Surplus = = 3 Summaries of Graphical Solution Methods TABLE 7.4 ISOPROFIT METHOD 1. Graph all s and find the feasible region. 2. Select a specific profit (or cost) line and graph it to find the slope. 3. Move the objective function line in the direction of increasing profit (or decreasing cost) while maintaining the slope. The last point it touches in the feasible region is the optimal solution. 4. Find the values of the decision variables at this last point and compute the profit (or cost). ORNER POINT METHOD 1. Graph all s and find the feasible region. 2. Find the corner points of the feasible reason. 3. ompute the profit (or cost) at each of the feasible corner points. 4. Select the corner point with the best value of the objective function found in Step 3. This is the optimal solution. opyright 2015 Pearson Education, Inc opyright 2015 Pearson Education, Inc Remember hapter 3: Where Prices ome From: The Interaction of Demand and Supply Read Quantitative Methods-module guide. Any questions please olivier.edu@gmail.com and make notes as you do so, in whatever way works best for you in terms of remembering information (your performance on this course is only assessed by exam). opyright 2010 Pearson Education, Inc. Economics R. Glenn Hubbard, Anthony Patrick O Brien, 3e. 39 of 46